If Cr $\equiv$ 25Cr and
C0 + 5.C1 + 9.C2 + .... + (101).C25 = 225.k, then k is equal to _____.
Answer (integer)
51
Solution
S = 1.<sup>25</sup>C<sub>0</sub> + 5.<sup>25</sup>C<sub>1</sub> + 9.<sup>25</sup>C<sub>2</sub> + .... + (101)<sup>25</sup>C<sub>25</sub>
<br><br>S = (101).<sup>25</sup>C<sub>25</sub> + (97).<sup>25</sup>C<sub>24</sub> + .......... + (1).<sup>25</sup>C<sub>0</sub>
<br>_________________________________________
<br><br>2S = 102{<sup>25</sup>C<sub>0</sub> + <sup>25</sup>C<sub>1</sub> + ......+ <sup>25</sup>C<sub>25</sub>}
<br><br>$\Rightarrow$ S = 51 $\times$ 2<sup>25</sup> = k.2<sup>25</sup>
<br><br>$\Rightarrow$ k = 51
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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